Question

$\underset{x\to 0}{\mathrm{Lim}}\frac{1}{\sin x}\underset{0}{\overset{\ln (1+x)}{\int }}(1-\tan 2y{)}^{1/y}dy$ equals

• A. $e^{-2}$
• B. $e$
• C. $e^2$
• D. $e^4$

Answer 1

Answer: (A)

$l=\underset{x\to 0}{\mathrm{Lim}}\frac{\underset{0}{\overset{\ln (1+x)}{\int }}(1-\tan 2y{)}^{1/y}dy}{\frac{\sin x}{x}\cdot x}$
Using L'Hospital's Rule
$l=\underset{x\to 0}{\mathrm{Lim}}\frac{[1-\tan 2(\ln (1+x)){]}^{\frac{1}{\ln (1+x)}}}{(1+x)}\left(1^{\infty }\right)$
$=e^{-\underset{x\to 0}{\mathrm{Lim}}\frac{1}{\ln (1+x)}\tan 2(\ln (1+x))}=e^{-\underset{x\to 0}{\mathrm{Lim}}\frac{\tan (2\ln (1+x))}{2\ln (1+x)}\cdot 2}=e^{-2}\left(\text{using}\underset{x\to 0}{\mathrm{Lim}}\frac{\tan \theta }{\theta }=1\right)$